Current going through resistor with Ohm's law?

Question

If the battery provides a 12 V electromotive force and the resistors are each 10 ohm , what is the current going through the resistor in region A in Amps?

I assume I need Ohm's law: V = IR.

I would think for branch A, I would plug in the values 12 = I * 10

because there is only one resistor in branch A. This gives me I = 1.2, which is incorrect.

Where am I going wrong?

Answer 1

Overall resistance in the circuit = $10 + \frac{10}2 + \frac{10}3 = \frac{55}3$ ohm.

Current $I = \frac VR = \frac{12}{\frac{55}3} = \frac{36}{55}$ ampere

Answer 2

Note that total resistance is equivalent to $R + (R^{-1} + R^{-1})^{-1} + (R^{-1} + R^{-1} + R^{-1})^{-1}$

Which means that $I_{a} = \frac{V}{R + (R^{-1} + R^{-1})^{-1} + (R^{-1} + R^{-1} + R^{-1})^{-1}}$

Answer 3

12V is dropped across the whole resistor network, not resistor $R_a$ (the resistor in region A). Less voltage is dropped across $R_a$.

Using Kirchoff's voltage law you can write:

$$V_a + V_b + V_c = 10$$

Using Kirchoff's current law you can write:

$$I_a = I_{batt}$$ $$I_a - I_{b1} - I_{b2} = 0$$ $$I_{b1}+I_{b2}-I_{c1}-I_{c2}-I_{c3}=0$$ $$I_{c1}+I_{c2}+I_{c3}=I_{batt}$$

And Ohm's law lets you write:

$$V_a=I_aR_a$$ $$V_{b}=I_{b1}R_{b1}$$ etc.

From that system of linear equations, and knowing that all the $R$'s are 10 ohms, you can solve for all the $V$'s and $I$'s.